MH
1
K
Khách
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22 tháng 10 2020
a) Ta có : 2005.2007 = (2006 - 1)(2006 + 1) = 20062 - 12 = 20062 - 1 ( cái khúc này sửa : 2005.2001 thành 2005.2007)
Mà B = 20062
=> 20062 - 1 < 20062
=> A < B
b) Ta có : B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (24 - 1)(24 + 1)(28 + 1)(216 + 1)
B = (28 - 1)(28 + 1)(216 + 1) = (216 - 1)(216 + 1) = 232 - 1
Mà C = 232
=> B < C
c) Tương tự như câu b
A
1
25 tháng 9 2021
a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
HA
2
12 tháng 9 2021
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
12 tháng 9 2021
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
NM
1
15 tháng 10 2023
\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\cdot\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
NH
1
DT
Dang Tung
CTVHS
8 tháng 11 2023
\(A=8.\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^8-1\right)....\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
KV
8 tháng 11 2023
A = 8.(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)
= (3¹⁶ - 1)(3¹⁶ + 1)
= 3³² - 1
27 tháng 6 2023
`A=4(3^2+1)(3^4+1)...(3^64+1)`
`=>2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`
- Ta có:
`(3^2-1)(3^2+1)=3^4-1`
`(3^4-1)(3^4+1)=3^16-1`
`....`
`(3^64-1)(3^64+1)=3^128-1`
Suy ra `2A=3^128-1=B`
`=>A<B`
C1: 2A = (3-1 )(3+1)....(3^16)
Theo hằng đẳng thức thì a^2-b^2 thì 2A = 3^32 - 1 => A = (3^32-1) /2<B
C2 biến đổi B
B = 3^32 - 1 = (3^16- 1)(3^16 + 1)=(3^8 - 1)(3^8 + 1)= ... = 2(3+1)(3^2+1)..(3^16+1)>A